The figure below shows three small, charged spheres, all lying along the horizontal axis. Sphere A, at left, has a 6.10 nC charge. Sphere B has a 1.60 nC charge and is 3.00 cm to the right of A. Sphere C has a
−2.50 nC
charge and is 2.00 cm to the right of B.
Find the magnitude (in N) and direction of the net electric force on each of the spheres.Three charges lie along a horizontal line. A positive charge labeled qA is on the left. 3.00 cm to its right is a positive charge labeled qB. 2.00 cm to the right of qB is a negative charge labeled qC.

To find the magnitude and direction of the net electric force on each of the charged spheres, we'll use Coulomb's law. Coulomb's law gives the force between two point charges as:

\[
F = k_e \frac{|q_1 q_2|}{r^2}
\]

Where:
- \( k_e \) is the Coulomb's constant, \( k_e = 8.99 \times 10^9 \ \text{N m}^2/\text{C}^2 \)
- \( q_1 \) and \( q_2 \) are the charges of the two spheres (in Coulombs)
- \( r \) is the distance between the two charges (in meters)

### Given:
- \( q_A = 6.10 \ \text{nC} = 6.10 \times 10^{-9} \ \text{C} \)
- \( q_B = 1.60 \ \text{nC} = 1.60 \times 10^{-9} \ \text{C} \)
- \( q_C = -2.50 \ \text{nC} = -2.50 \times 10^{-9} \ \text{C} \)
- Distance between A and B: \( r_{AB} = 3.00 \ \text{cm} = 0.0300 \ \text{m} \)
- Distance between B and C: \( r_{BC} = 2.00 \ \text{cm} = 0.0200 \ \text{m} \)
- Distance between A and C: \( r_{AC} = r_{AB} + r_{BC} = 0.0500 \ \text{m} \)

### Steps to Calculate Forces:
1. **Force on B due to A** (\( F_{AB} \)):
   \[
   F_{AB} = k_e \frac{|q_A \cdot q_B|}{r_{AB}^2}
   \]

2. **Force on B due to C** (\( F_{BC} \)):
   \[
   F_{BC} = k_e \frac{|q_B \cdot q_C|}{r_{BC}^2}
   \]

3. **Force on A due to B** (\( F_{AB} \)), same as calculated above, but with different direction.

4. **Force on A due to C** (\( F_{AC} \)):
   \[
   F_{AC} = k_e \frac{|q_A \cdot q_C|}{r_{AC}^2}
   \]

5. **Force on C due to B** (\( F_{CB} \)), same as \( F_{BC} \) but direction changes.

6. **Force on C due to A** (\( F_{CA} \)), same as \( F_{AC} \).

Next, calculate each force and sum them up considering their directions.

Let’s compute each force:

### Step 1: Calculate the force on B due to A (\( F_{AB} \)):
\[
F_{AB} = 8.99 \times 10^9 \times \frac{(6.10 \times 10^{-9})(1.60 \times 10^{-9})}{(0.0300)^2} 
\]

### Step 2: Calculate the force on B due to C (\( F_{BC} \)):
\[
F_{BC} = 8.99 \times 10^9 \times \frac{(1.60 \times 10^{-9})(2.50 \times 10^{-9})}{(0.0200)^2}
\]

### Step 3: Calculate force on A due to C (\( F_{AC} \)):
\[
F_{AC} = 8.99 \times 10^9 \times \frac{(6.10 \times 10^{-9})(2.50 \times 10^{-9})}{(0.0500)^2}
\]

After calculating these, sum the forces considering directions and compute the net force on each sphere.

Let me do the calculations.The calculated forces are:

1. **Force on B due to A (\( F_{AB} \))**: \( 9.75 \times 10^{-5} \ \text{N} \) to the right.
2. **Force on B due to C (\( F_{BC} \))**: \( 8.99 \times 10^{-5} \ \text{N} \) to the left.
3. **Force on A due to C (\( F_{AC} \))**: \( 5.48 \times 10^{-5} \ \text{N} \) to the left.

### Net Force on Each Sphere:
1. **Sphere A**: The net force on A is due to B (right) and C (left).
2. **Sphere B**: The net force on B is the sum of the forces from A (right) and C (left).
3. **Sphere C**: The net force on C is due to A (right) and B (left).

Let’s summarize the net forces and directions using the vector sum of these forces.

Learn how to calculate the net electric force on three charged spheres using Coulomb's law. Detailed step-by-step solution for spheres with charges 6.10 nC, 1.60 nC, and -2.50 nC spaced along a horizontal axis. Suitable for college-level physics students.

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